Math 10a
Practice Midterm 2 #2
1. In summation notation, write down the left Riemann sum estimate for
using 1000 intervals.
999
1 X k
1000 k=0 1000
k
1−
1000
R1
0
x(1 − x)dx
.
2. (a) What is the Taylor series for ln(x) centered at x = 1?
Differentiating f (x) = ln(x):
1
f 0 (x) =
x
1
f 00 (x) = − 2
x
2
f 000 (x) = 3
x
3·2
f (4) (x) = − 4
x
4·3·2
.
f (5) (x) = −
x5
We note a pattern
(k − 1)!
f (k) (x) = (−1)k+1
, k ≥ 1.
xk
f (k) (1) = (−1)k+1 (k − 1)!, k ≥ 1.
Hence
f (x) =
∞
X
k=0
f
(k)
∞
∞
X
(x − 1)k
(−1)k+1 (k − 1)!(x − 1)k X (−1)k+1 (x − 1)k
(1)
= ln(1)+
=
.
k!
k!
k
k=1
k=1
Note that we have to be careful to split off the k = 0 term, since our formula for
the derivative doesn’t work for k = 0.
(b) What is the radius of the convergence of the series from part (a)?
I see factorials, so I perform the ratio test:
k+2 (x−1)k+1 k(x − 1) ak+1 (−1) k+1
=
ak (−1)k+1 (x−1)k = k + 1 → |x − 1|
k
so the series converges for |x − 1| < 1, diverges for |x − 1| > 1, so the radius of
convergence is 1.
1
(c) Write down a series of rational numbers converging to ln(1/3).
Since the series coverges for x = 31 ,
k
X
∞
∞
X
(−1)k+1 31 − 1
1
2k
ln
=
=
− k .
3
k
3 k
k=1
k=1
3. (a) Suppose giraffe neck lengths are normally distributed with mean of 6 feet and a
standard deviation of 6 inches. What is the probability, given a randomly selected
giraffe, that its neck is shorter than 5 feet?
5 feet is two standard deviations below the mean, and the area under the bell
curve to the left of −2 standard deviations is .025. Hence the answer is 2.5%.
(b) Suppose giraffe tongue lengths are normally distributed with a mean of 20 inches
(!) and a standard deviation of 3 inches. What is the probability that a randomly
selected giraffe will have a tongue of length between 20 and 23 inches?
20 is the mean and 23 is one standard deviation above the mean. Hence we’re
looking at the area under a bell curve between the mean and one standard deviation above the mean. This is half the area between ±1 standard deviations,
hence is 34%.
4. Compute the following integrals:
(a)
Z
x
dx
1−x
u = 1 − x, du = −dx, x = 1 − u
Z Z
1
1−u
(−du) =
1−
du = u − ln |u| + C = 1 − x − ln |1 − x| + C
u
u
(b)
Z
√
x x + 1dx
u = x + 1, du = dx, x = u − 1
Z
Z
√
u5/2 u3/2
2
2
(u − 1) udu = u3/2 − u1/2 =
−
= (x + 1)5/2 − (x + 1)3/2 + C.
5/2
3/2
5
3
(c)
Z
See the lecture slides.
2
ex sin(x)dx
(d)
√
sin( x)dx
Z
u=
√
x, du =
1 √1
dx
2 x
dx
2u
=
⇒ dx = 2udu
Z
=2
Z
= −2u cos(u)+2
Z
u sin(u)du = 2
u
d
(− cos(u))du
du
√
√
√
cos(u) = −2u cos(u)+2 sin(u)+C = −2 x cos( x)+2 sin( x)+C.
5. Compute the following integrals:
(a)
2
Z
2
u = 1 + x , du =
2xdx 12
R x=2
du
√
x=1
u
x
√
dx
1 + x2
1
R u=5
= 12 u=2 √duu =
(b)
Z
u=5
1 u1/2 2 1/2 =
√
5−
√
2.
u=2
π
x sin(x)dx
0
Z
π
x
0
π
Z π
d
− cos(x) dx = −x cos(x) + 2
cos(x)dx = π.
dx
0
0
6. Recall that the uniform distribution from 0 to 1 is defined to be one whose pdf is
(
1 x ∈ [0, 1]
f (x) =
.
0 otherwise
What is the cdf of this uniform distribution? Sketch a graph.
I’ll leave it to you to sketch the graph. The cdf F (x) is the area under the pdf to the
left of x, so

Z x
Z x 
0 t < 0
F (x) =
f (t)dt =
1 0 ≤ t ≤ 1 dt
−∞
−∞ 

0 t>1
R x



x<0
R−∞ 0dt R
0 x < 0
0
x
=
0dt + 0 1dt
0≤x≤1 = x 0≤x≤1.
−∞


R1
Rx
R 0

0dt + 0 1dt + 1 0dt x > 1
1 x>1
−∞
3